Step-by-Step Solution

Topic: Types of Functions (One-One & Onto)

Visual Representation: Mapping Diagram

The diagram illustrates the mapping of inputs (Domain) to outputs (Codomain) for f(x) = x².

(I) RANGE OF f: N → R

The domain is N = {1, 2, 3, 4, …}. The function is f(x) = x².

• • f(1) = 1² = 1
• • f(2) = 2² = 4
• • f(3) = 3² = 9

The Range is the set of all output values (Perfect Squares).

(II) CHECK FOR INJECTIVE (ONE-ONE)

A function is injective if f(x₁) = f(x₂) implies x₁ = x₂.

Let x₁, x₂ ∈ N (Natural Numbers).

If x₁² = x₂², then x₁ = ±x₂.

Since the domain is N, x cannot be negative. Thus, x₁ = x₂.

(III)(A) PROVE BIJECTIVE

1. Injective (One-One): As proved in (ii), distinct natural numbers have distinct squares.

2. Surjective (Onto):

Codomain = {1, 4, 9, 16, …} which is exactly the set of perfect squares.

For every element ‘y’ in the codomain, there exists an element ‘x’ = √y in the domain such that f(x) = y.

Since Range = Codomain, it is Surjective.

(III)(B) NEITHER INJECTIVE NOR SURJECTIVE (Real Numbers)

Domain = R, Codomain = R.

Check Injective:

Take x = 1 and x = -1.

f(1) = 1² = 1

f(-1) = (-1)² = 1

Since different inputs give the same output (1), it is NOT Injective.

Check Surjective:

The output x² is always non-negative (x² ≥ 0).

Negative numbers in the Codomain (like -2, -5) have no pre-image in the Domain.

Range [0, ∞) ≠ Codomain R. Thus, it is NOT Surjective.